<div dir="ltr"><div>Hello Avraham --Happy to see you</div><div><br></div><div>My code is trying to produce a cube c which is going to be constructed by a vector a and matrix b.</div><div>And the number of rows in b and size of a must be same.</div><div><br></div><div>So we can assume that if a is a vector of size 3, Then b must be 2 x 3 or 3 X 3 etc. </div><div><br></div><div>Thank you Avraham for quick response. I hope this will make my question more clear.</div><div><br></div><div>Best regards</div><div><br></div></div><div class="gmail_extra"><br><div class="gmail_quote">On Wed, Dec 14, 2016 at 4:46 PM, Avraham Adler <span dir="ltr"><<a href="mailto:avraham.adler@gmail.com" target="_blank">avraham.adler@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div class="gmail_extra"><div class="gmail_quote"><div><div class="h5">On Tue, Dec 13, 2016 at 9:51 PM, Amina Shahzadi <span dir="ltr"><<a href="mailto:aminashahzadi@gmail.com" target="_blank">aminashahzadi@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="ltr"><div><div><div>Hello Friends and Prof. Dirk<br><br></div>I am pasting here a code which has a for loop depending on another for loop. <br></div>I am getting zeros for cube c. I tried and searched a lot but did not get an example of this type. Would you please help in this regard?<br><br><br>#include <RcppArmadillo.h><br>using namespace Rcpp;<br>using namespace RcppArmadillo;<br>using namespace arma;<br>//[[Rcpp::depends(RcppArmadill<wbr>o)]]<br>//[[Rcpp::export]]<br><br><br>arma::cube exam(arma::vec a, arma::mat b)<br>{<br> int m = a.size();<br> int n = b.n_rows;<br> arma::cube c = zeros<cube>(n, m, n);<br> for(int i=0; i<n; i++) {<br> for(int j=0; j<m; j++) {<br> for(int k=0; k<i; k++) {<br> if(k==0) {<br> c(i, j ,k) = c(i, j, k) + b(i, j);<br> }<br> else{<br> c(i, j, k) = c(i, j, k) + c(i-1, j, k) *b(i, j);<br> }<br> }<br> }<br> }<br> return c;<br>}<br><br><br></div><div><div><div><div>Thank You<span class="m_-460033997963071725gmail-HOEnZb"><font color="#888888"><br></font></span></div><span class="m_-460033997963071725gmail-HOEnZb"><font color="#888888"><div>-- <br><div class="m_-460033997963071725gmail-m_-5403746280849366164gmail_signature"><i><font face="comic sans ms,sans-serif">Amina Shahzadi</font></i></div>
</div></font></span></div></div></div></div></blockquote></div></div><div><br><br><div>Hello.<br><br></div><div>I haven't run your code, but it strikes me
that I cannot see where are you capturing the number of columns of b.
It's a bit confusing as I was always taught a matrix has m rows and n
columns. Be that as it may, your k==0 loop looks like it's trying to
copy over the original matrix to the first slice, but how do you know
that b has m columns, which is what you're assuming by letting j loop to
m. Unless you are assuming a square matrix?<br><br></div><div>Even if you are, if your matrix is not the same length as your vector, I think there is an issue with your loop boundaries, unless I've misunderstood something, which is certainly possible.<br><br>For example, assume a is {1, 2, 3} and b is the 2 x 2 of row 1: [1 2] and row 2: [3 4]. Thus m = 3 and n = 2.<br><br></div><div>Step 1: i = j = k = 0: c(0, 0, 0) becomes b(0, 0) or 1.<br><br></div><div>Step 2: i = 0, j = 1, k = 0: c(0, 1, 0) becomes b(0, 1) or 2.<br><br></div><div>Step 3: i = 0, j = 2, k = 0: c(0, 2, 0) becomes b(0, 2) ?!?! There is no b(0, 2), it's only a 2x2 matrix?<br><br><br></div><div>Similar to your previous questions, instead of posting code, can you please describe in words what it is you are trying to do? That may help.<br></div><div><br></div>Avi <br></div></div><br></div></div>
</blockquote></div><br><br clear="all"><br>-- <br><div class="gmail_signature" data-smartmail="gmail_signature"><em><font face="comic sans ms,sans-serif">Amina Shahzadi</font></em></div>
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